Unable to narrow parameterized generic types for functions

[lingz]

TypeScript Version: 2.8

Code

function test<T extends 'a' | 'b'>(x: T) {
  if (x === 'a') {
    const z1 = x;
  }
}

function test2(x: 'a' | 'b') {
  if (x === 'a') {
    const z2 = x;
  }
}

Expected behavior:
z1 and z2 should have the same type, which is a

Actual behavior:
function test is unable to narrow parameter x, with the if check, and z2 resolves to x

[MartinJohns]

With strictNullChecks enabled your code won't even compile, because the type null is not compatible with the type 'a' | 'b'.

Without strictNullChecks the value null is implicitly compatible with every type. When you call test(null) the compiler resolves to any because it can't figure out the type of T. Based on null the compiler can't deduct any sensible type. You could expect that it should return 'a' | 'b', because T must extend from it, but that would still only be a "guess". There's no T-type known.

[lingz]

I think I communicated the issue confusingly, it was not about the actual return type but the type guard in the implementation, I edited the first post to reflect that.

[mhegazy]

Duplicate of #13995

[typescript-bot]

Automatically closing this issue for housekeeping purposes. The issue labels indicate that it is unactionable at the moment or has already been addressed.